Problem: What is the extraneous solution to these equations? $\dfrac{x^2 + 20}{x + 8} = \dfrac{-4x + 52}{x + 8}$
Explanation: Multiply both sides by $x + 8$ $ \dfrac{x^2 + 20}{x + 8} (x + 8) = \dfrac{-4x + 52}{x + 8} (x + 8)$ $ x^2 + 20 = -4x + 52$ Subtract $-4x + 52$ from both sides: $ x^2 + 20 - (-4x + 52) = -4x + 52 - (-4x + 52)$ $ x^2 + 20 + 4x - 52 = 0$ $ x^2 - 32 + 4x = 0$ Factor the expression: $ (x - 4)(x + 8) = 0$ Therefore $x = 4$ or $x = -8$ At $x = -8$ , the denominator of the original expression is 0. Since the expression is undefined at $x = -8$, it is an extraneous solution.